Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.ArrayList;

import java.util.Iterator;
import java.util.Scanner;



public class main {

public static void main(String[] args) {

Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int k=sc.nextInt();
int arr[][]=new int[n][k];
int arrd[][]=new int[n][k];

int arr1[]=new int[n];
int arr2[]=new int[k];
int arr1d[]=new int[n];
int arr2d[]=new int[k];

for(int i=0;i<n;i++){
	for(int j=0;j<k;j++){
		arr[i][j]=sc.nextInt();
		arrd[i][j]=arr[i][j];
		
		
	}
}

ArrayList<String> al=new ArrayList();

int count=0;

for(int i=0;i<k;i++){
	int min=Integer.MAX_VALUE;
	for(int j=0;j<n;j++){
		if(arr[j][i]<min) min=arr[j][i];
	}
	arr2[i]=min;
}

for(int i=0;i<k;i++){
	int a=arr2[i];
	if(a>0){ 	for(int j=0;j<n;j++){
		arr[j][i]=arr[j][i]-a;
	}}
	while(a!=0){
		count=count+1;
		al.add("col "+(i+1));
		a--;
	}

}

for(int i=0;i<n;i++){
	int min=Integer.MAX_VALUE;
	for(int j=0;j<k;j++){
		if(arr[i][j]<min) min=arr[i][j];
	}
	arr1[i]=min;
}

for(int i=0;i<n;i++){
	int a=arr1[i];
	if(a>0){ 
	for(int j=0;j<k;j++){
		arr[i][j]=arr[i][j]-a;
	}}
	while(a!=0){
		count=count+1;
		al.add("row "+(i+1));
		a--;
	}
	
}

int comp=1;
for(int i=0;i<n;i++){
	for(int j=0;j<k;j++){
		if(arr[i][j]!=0){
			comp=0;
			break;
		}
	}
}













int count1=0;
ArrayList<String> al1=new ArrayList(); 
for(int i=0;i<n;i++){
	int min=Integer.MAX_VALUE;
	for(int j=0;j<k;j++){
		if(arrd[i][j]<min) min=arrd[i][j];
	}
	arr1d[i]=min;
}

for(int i=0;i<n;i++){
	int a=arr1d[i];
	if(a>0){ 
	for(int j=0;j<k;j++){
		arrd[i][j]=arrd[i][j]-a;
	}}
	while(a!=0){
		count1=count1+1;
		al1.add("row "+(i+1));
		a--;
	}
	
}
for(int i=0;i<k;i++){
	int min=Integer.MAX_VALUE;
	for(int j=0;j<n;j++){
		if(arrd[j][i]<min) min=arrd[j][i];
	}
	arr2d[i]=min;
}

for(int i=0;i<k;i++){
	int a=arr2d[i];
	if(a>0){ 	for(int j=0;j<n;j++){
		arrd[j][i]=arrd[j][i]-a;
	}}
	while(a!=0){
		count1=count1+1;
		al1.add("col "+(i+1));
		a--;
	}

}
int comp1=1;
for(int i=0;i<n;i++){
	for(int j=0;j<k;j++){
		if(arr[i][j]!=0){
			comp1=0;
			if(comp==0){
				System.out.println(-1); return;
			}
		}
	}
}

if((count<count1&&comp!=0)||comp1==0){
System.out.println(count);
Iterator it =al.iterator();
while(it.hasNext()){
	System.out.println(it.next());
}
}
else{
	
	System.out.println(count1);
	Iterator it =al1.iterator();
	while(it.hasNext()){
		System.out.println(it.next());
	}
	
}
}


}